# Quotient by subgroup is well defined if and only if subgroup is normal

Let $$G$$ be a group and $$N$$ a nor­mal sub­group of $$G$$. Then we may define the quo­tient group $$G/N$$ to be the set of left cosets $$gN$$ of $$N$$ in $$G$$, with the group op­er­a­tion that $$gN + hN = (gh)N$$. This is well-defined if and only if $$N$$ is nor­mal.

# Proof

## $$N$$ nor­mal im­plies $$G/N$$ well-defined

Re­call that $$G/N$$ is well-defined if “it doesn’t mat­ter which way we rep­re­sent a coset”: whichever coset rep­re­sen­ta­tives we use, we get the same an­swer.

Sup­pose $$N$$ is a nor­mal sub­group of $$G$$. We need to show that given two rep­re­sen­ta­tives $$g_1 N = g_2 N$$ of a coset, and given rep­re­sen­ta­tives $$h_1 N = h_2 N$$ of an­other coset, that $$(g_1 h_1) N = (g_2 h_2)N$$.

So given an el­e­ment of $$g_1 h_1 N$$, we need to show it is in $$g_2 h_2 N$$, and vice versa.

Let $$g_1 h_1 n \in g_1 h_1 N$$; we need to show that $$h_2^{-1} g_2^{-1} g_1 h_1 n \in N$$, or equiv­a­lently that $$h_2^{-1} g_2^{-1} g_1 h_1 \in N$$.

But $$g_2^{-1} g_1 \in N$$ be­cause $$g_1 N = g_2 N$$; let $$g_2^{-1} g_1 = m$$. Similarly $$h_2^{-1} h_1 \in N$$ be­cause $$h_1 N = h_2 N$$; let $$h_2^{-1} h_1 = p$$.

Then we need to show that $$h_2^{-1} m h_1 \in N$$, or equiv­a­lently that $$p h_1^{-1} m h_1 \in N$$.

Since $$N$$ is closed un­der con­ju­ga­tion and $$m \in N$$, we must have that $$h_1^{-1} m h_1 \in N$$; and since $$p \in N$$ and $$N$$ is closed un­der mul­ti­pli­ca­tion, we must have $$p h_1^{-1} m h_1 \in N$$ as re­quired.

## $$G/N$$ well-defined im­plies $$N$$ normal

Fix $$h \in G$$, and con­sider $$hnh^{-1} N + hN$$. Since the quo­tient is well-defined, this is $$(hnh^{-1}h) N$$, which is $$hnN$$ or $$hN$$ (since $$nN = N$$, be­cause $$N$$ is a sub­group of $$G$$ and hence is closed un­der the group op­er­a­tion). But that means $$hnh^{-1}N$$ is the iden­tity el­e­ment of the quo­tient group, since when we added it to $$hN$$ we ob­tained $$hN$$ it­self.

That is, $$hnh^{-1}N = N$$. There­fore $$hnh^{-1} \in N$$.

Since this rea­son­ing works for any $$h \in G$$, it fol­lows that $$N$$ is closed un­der con­ju­ga­tion by el­e­ments of $$G$$, and hence is nor­mal.

Parents:

• Normal subgroup

Nor­mal sub­groups are sub­groups which are in some sense “the same from all points of view”.